This is the same bonus problem as Lesson 12. This time, you have to use the method of Lagrange Multipliers to solve it. A rectangular tank with a bottom and sides but no top is to have volume 500 cubic feet. Determine the dimensions (length, width, height) with the smallest possible surface area.

length=10 ft, width = 10 ft and height = 5 ft

Step-by-step explanation:

using Lagrange multipliers, we have the main function that is the Area A of the tank,

A (x, y, z) = x*y + 2*x*z + 2*y*z

constrained to the Volume V (x, y, z) = x*y*z=a = 500 ft³

using Lagrange multipliers

Ax - λ*Vx = 0 → (y + 2*z) - y*z*λ = 0 → λ = 1/z + 2/y

Ay - λ*Vy = 0 → (x + 2*z) - x*z*λ = 0 → λ = 1/z + 2/x

Az - λ*Vz = 0 → (2*x + 2*y) - x*y*λ = 0 → λ = 2/x + 2/y

V = a → x*y*z=a

adding the first and second equations

2*λ = 1/z + 2/y + 1/z + 2/x = 2/z + λ

λ = 2/z → z = 2/λ

therefore

λ = 1/z + 2/y = λ/2 + 2/y

λ/2 = 2/y → y = 4/λ

and similarly x=4/λ

then

x*y*z=a

2/λ*4/λ*4/λ = a

32/λ³ = a

λ = ∛32/a

therefore

z = 2/λ = 2*∛a/32 = 2*∛ (500/32) = 5

y = 4/λ = 2*5 = 10

x=10

therefore the dimensions that minimize the area are x=10, y=10 and z=5 (length=10 ft, width = 10 ft and height = 5 ft)